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stepan [7]
3 years ago
13

In this experiment, the

Physics
1 answer:
tigry1 [53]3 years ago
6 0
Result of the other variable
You might be interested in
A certain heat engine takes in 300 J of energy from a hot source and then transfers 200 J of that energy to a colder object. Wha
Greeley [361]

Answer:

The efficiency is 0.33, or 33%.

Explanation:

From the thermodynamics equations, we know that the formula for the efficiency of a heat engine is:

\eta=1-\frac{Q_2}{Q_1}

Where η is the efficiency of the engine, Q_1 is the heat energy taken from the hot source and Q_2 is the heat energy given to the cold object. So, plugging the given values in the formula, we obtain:

\eta=1-\frac{200J}{300J}=0.33

This means that the efficiency of the heat engine is 0.33, or 33% (The efficiency of an engine is dimensionless).

5 0
3 years ago
The force that pulls away from a curve is known as __________ .
tatyana61 [14]

Answer: <u>Centrifugal Force </u>

Explanation:

The force that pulls out from the center on a body in circular motion is called centrifugal force and it increases with acceleration. Centrifugal force results in strong outward pull on your vehicle.

4 0
1 year ago
Let's talk skateboard velocity how much mean when you get my going down a 3 x 3" metal ramp. No posers are allowed to answer th
denpristay [2]

Answer:

quarte embri Oslo

Explanation:

8 0
3 years ago
Read 2 more answers
A series RCL circuit is at resonance and contains a variable resistor that is set to 206Ω. The power dissipated in the circuit i
mash [69]

Answer:

Power dissipated in resistor 532 ohm is 0.503 watt

Explanation:

We have given in first case resistance R_1=206ohm

Power dissipated in this resistance is P_1=1.30watt

Power dissipated in the resistor is equal to P=\frac{v_{rms}}^2{R}

We have to find the power dissipated in the resistor is 1.30 watt

From the relation we can say that \frac{P_1}{P_2}=\frac{R_2}{R_1}

\frac{1.3}{P_2}=\frac{532}{206}

P_2=0.503watt

So power dissipated in resistor 532 ohm is 0.503 watt  

5 0
3 years ago
A myopic person (assume no astigmatism) is diagnosed with a far point of 160 cm. What corrective prescription should they be sup
Kazeer [188]

Answer:

The person should use a convex lens of power -0.625 D.

Explanation:

Given that, the far point of the person, with myopic eye, is 160 cm.

It means that the image of an object, which is at infinity, is formed at this point 160 cm distance away from the person's eye.

If the object is infinity, then the object distance, u = -\infty.

Image distance, v = -160 cm.

u and v are taken to be negative because the object and the image of the object are on the side of the eye from where the light is coming.

Let the focal length of the lens which is to be used for the correction of this myopic eye be f, then using lens equation,

\rm \dfrac 1f = \dfrac 1v-\dfrac 1u=\dfrac 1{-160 }-\dfrac1{-\infty}=\dfrac 1{-160}-0=\dfrac 1{-160}.\\\Rightarrow f=-160\ cm.

The negative focal length indicates that the lens should be convex.

The power of the lens is given by

\rm P=\dfrac{1}{f}=\dfrac{1}{-160\ cm }=\dfrac{1}{-1.6\ m}=-0.625\ D.

So, the person should use a convex lens of power -0.625 D.

8 0
4 years ago
Read 2 more answers
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